1. ## Vectors in 3D

I started solving this by taking the first point as x1,y1,z1 and second one as x2,y2,z2
and so the distance between these points to B is *sqrt*14.
Although this method doesn't seem to be working. Any help/hint on how to proceed with this question will be much appreciated.

2. ## Re: Vectors in 3D

You need to find where the y-axis insects the sphere $(x+1)^2+(y+1)^2+(z-2)^2=14~.$
Recall that the y-axis is $(0,t,0)~.$

3. ## Re: Vectors in 3D

So i put y=t? and x=0 and z=0?

4. ## Re: Vectors in 3D

Originally Posted by kandyfloss
So i put y=t? and x=0 and z=0?
Indeed, yes. Solve for t.

5. ## Re: Vectors in 3D

Originally Posted by Plato
Indeed, yes. Solve for t.
So i get t(^2) + 2t -13=0
And it seems my values for t are going to be in fraction, and the answer is in whole numbers.

6. ## Re: Vectors in 3D

Originally Posted by kandyfloss
So i get t(^2) + 2t -13=0
I get $t^2+2t-8=0$

7. ## Re: Vectors in 3D

Originally Posted by Plato
I get $t^2+2t-8=0$
oh yes, sorry. I'm kind of out of my mind today. Thanks a lot!